Tính:
\(\frac{2\sqrt{3}-1}{\sqrt{15}}-\frac{2-\sqrt{5}}{\sqrt{3}}-\frac{4\sqrt{15}-10\sqrt{3}}{15}\)
Thực hiện phép tính
1)\(\frac{\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}+\sqrt{2}}{\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}+\sqrt{5}}\)
2)\(\left(4+\sqrt{15}\right)\left(10-\sqrt{6}\right)-\sqrt{4-\sqrt{15}}\)
3)\(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
4)\(\frac{2\sqrt{3-\sqrt{5+\sqrt{13-\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
5)\(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
Rút gọn biểu thức
1)\(\frac{15}{3\sqrt{20}}\)
2) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{2}-\sqrt{5}}\)
3) \(\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{6}+\sqrt{2}}\)
4) \(\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}\)
5) \(\left(\sqrt{20}-\sqrt{45}+\sqrt{5}\right)\sqrt{5}\)
6)\(\left(2+\sqrt{5}\right)^2-\left(2+\sqrt{5}\right)^2\)
7) \(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\right):2\sqrt{5}\)
8)\(\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}\)
9) \(2\sqrt{3}\left(2\sqrt{6}-\sqrt{3}+1\right)\)
10) \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
11) \(\sqrt{\sqrt{10}+1}.\sqrt{\sqrt{10}-1}\)
12) \(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
13) \(\sqrt{\frac{3}{4}}+\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{12}}\)
14) \(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}\right)\sqrt{6}\)
15 ) \(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)
16) \(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
17) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
tính:
a)\(\frac{1}{1+\sqrt{5}}+\frac{1}{1-\sqrt{5}}\)
b)\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}\)
c)\(\frac{2}{\sqrt{5}+1}+\sqrt{\frac{2}{3-\sqrt{5}}}-5\sqrt{\frac{1}{5}}\)
d)\(\left(\frac{5}{\sqrt{15}-\sqrt{10}}-\frac{3\sqrt{5}-5\sqrt{3}}{\sqrt{3}-\sqrt{5}}\right)^2\)
e)\(\frac{2}{\sqrt{3}-\sqrt{5}}+\frac{3-2\sqrt{3}}{\sqrt{3}-2}\)
Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
Tính
a) \(\frac{10}{\sqrt{5}}+\frac{8}{3 +\sqrt{5}}-\frac{\sqrt{15}-2\sqrt{5}}{\sqrt{3}-2}\)
b) \(\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}-\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{5}-\sqrt{7}}\)
Rút gọn
a, \(\frac{2\sqrt{3-1}}{\sqrt{15}}-\frac{2-\sqrt{5}}{\sqrt{3}}-\frac{4\sqrt{15}-10\sqrt{3}}{15}\)
b, \(\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\left(\frac{\sqrt{a}+1}{\sqrt{a-1}}+\frac{\sqrt{a-1}}{\sqrt{a}+1}\right)\)
c, \(\sqrt{4+\sqrt{7}-\sqrt{4-\sqrt{7}}}\)
d, \(6+2\sqrt{2}.3-\sqrt{4+\sqrt{2\sqrt{3}}}\)
e, \(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
Help me !!!
Tính
\(\frac{\sqrt{10}+5\sqrt{3}}{\sqrt{15}+\sqrt{5}}-\frac{3}{2\sqrt{2}-\sqrt{5}}+\sqrt{9+4\sqrt{2}}\)
thực hiện phép tính:
a)\(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}\)
b)\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
c)\(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
a) Kết quả rút gọn xấu (+dài) nữa. (có thể đề sai)
b)
\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left[\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right].\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\)
c) \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{\left(\sqrt{5}-\sqrt{2}\right)^2}{3}\)
a) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}=\left[\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right].\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}=\frac{1}{2}-2=-\frac{3}{2}\)
\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\\ \\ \\ \sqrt{\frac{9}{4}-\sqrt{2}}\\ \\ \\ Sosanh2\sqrt{27}va\sqrt{147}\\ \\ \\ 2\sqrt{15}va\sqrt{59}\\ \\ \\ 2\sqrt{2}-1va2\\ \\ \\ \frac{\sqrt{3}}{2}va1\\ \\ \\ -\frac{\sqrt{10}}{2}va-2\sqrt{5}\\ \\ \\ \sqrt{6}-1va3\\ \\ \\ 2\sqrt{5}-5\sqrt{2}va1\\ \\ \\ \frac{\sqrt{8}}{3}va\frac{3}{4}\\ \\ \\ -2\sqrt{6}va-\sqrt{23}\\ \\ \\ 2\sqrt{6}-2va3\\ \\ \\ \sqrt{111}-7va4\)
Xếp theo thứ tự tăng dần: \(21,2\sqrt{7},15\sqrt{3},-\sqrt{123}\) ; \(28\sqrt{2},\sqrt{14},2\sqrt{147},36\sqrt{4}\)
giảm dần: \(6\sqrt{\frac{1}{4}},4\sqrt{\frac{1}{2}},-\sqrt{132},2\sqrt{3},\sqrt{\frac{15}{5}}\); \(-27,4\sqrt{3},16\sqrt{5},21\sqrt{2}\)
a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)
=-7
b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)
So sánh:
1) \(2\sqrt{27}\) và \(\sqrt{147}\)
+ \(2\sqrt{27}\) = \(6\sqrt{3}\)
+ \(\sqrt{147}\) = \(7\sqrt{3}\)
⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)
Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)
2) \(2\sqrt{15}\) và \(\sqrt{59}\)
+ \(2\sqrt{15}\) = \(\sqrt{60}\)
⇒ \(\sqrt{60}\) > \(\sqrt{59}\)
Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)
3) \(2\sqrt{2}-1\) và 2
\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)
So sánh: 3 và \(2\sqrt{2}\)
+ 3 = \(\sqrt{9}\)
+ \(2\sqrt{2}=\sqrt{8}\)
⇒ \(\sqrt{8}\) < \(\sqrt{9}\)
⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1
⇒ \(2\sqrt{2}\) - 1 < 3 - 1
Vậy: \(2\sqrt{2}-1< 2\)
4) \(\frac{\sqrt{3}}{2}\) và 1
+ 1 = \(\frac{2}{2}\)
⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)
Vậy: \(\frac{\sqrt{3}}{2}\) < 1
5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)
+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)
⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)
Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)